Combination Formula

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The formula for combinations, also known as binomial coefficients, is represented as nCr, where n is the total number of objects and r is the number of objects to be chosen. The formula for nCr is: nCr = n! / (r! * (n-r)!)

The combination formula is used to find the number of ways of selecting items from a collection, such that the order of selection does not matter. In simple words, combination involves the selection of objects or things out of a larger group where order doesn’t matter.

The formula for combination helps to find the number of possible combinations that can be obtained by taking a subset of items from a larger set. It shows how many different possible subsets can be made from the larger set. It should be noted that the formula for permutation and combination are interrelated and are mentioned below.

Formula for Combination

Notations in nCr Formula:

  • r is the size of each permutation
  • n is the size of the set from which elements are permuted
  • n, r are non-negative integers
  • ! is the factorial operator

The combination formula shows the number of ways a sample of “r” elements can be obtained from a larger set of “n” distinguishable objects.

Example Question From Combination Formula

Question 1: Father asks his son to choose 4 items from the table. If the table has 18 items to choose, how many different answers could the son give?

Solution:
Given,
r = 4 (item sub-set)
n = 18 (larger item)

Therefore, simply: find “18 Choose 4”

We know that, Combination = C(n, r) = n!/r!(n–r)!

Combination Problem 1

Choose 2 Prizes from a Set of 6 Prizes

You have won first place in a contest and are allowed to choose 2 prizes from a table that has 6 prizes numbered 1 through 6. How many different combinations of 2 prizes could you possibly choose?

In this example, we are taking a subset of 2 prizes (r) from a larger set of 6 prizes (n). Looking at the formula, we must calculate “6 choose 2.”

C (6,2)= 6!/(2! * (6-2)!) = 6!/(2! * 4!) = 15 Possible Prize Combinations

The 15 potential combinations are {1,2}, {1,3}, {1,4}, {1,5}, {1,6}, {2,3}, {2,4}, {2,5}, {2,6}, {3,4}, {3,5}, {3,6}, {4,5}, {4,6}, {5,6}

Combination Problem 2

Choose 3 Students from a Class of 25

A teacher is going to choose 3 students from her class to compete in the spelling bee. She wants to figure out how many unique teams of 3 can be created from her class of 25.

In this example, we are taking a subset of 3 students (r) from a larger set of 25 students (n). Looking at the formula, we must calculate “25 choose 3.”

C (25,3)= 25!/(3! * (25-3)!)= 2,300 Possible Teams

Combination Problem 3

Choose 4 Menu Items from a Menu of 18 Items

A restaurant asks some of its frequent customers to choose their favorite 4 items on the menu. If the menu has 18 items to choose from, how many different answers could the customers give?

Here we take a 4 item subset (r) from the larger 18 item menu (n). Therefore, we must simply find “18 choose 4.”

C (18,4)= 18!/(4! * (18-4)!)= 3,060 Possible Answers

Handshake Problem

In a group of n people, how many different handshakes are possible?

First, let’s find the total handshakes that are possible. That is to say, if each person shook hands once with every other person in the group, what is the total number of handshakes that occur?

A way of considering this is that each person in the group will make a total of n-1 handshakes. Since there are n people, there would be n times (n-1) total handshakes. In other words, the total number of people multiplied by the number of handshakes that each can make will be the total handshakes. A group of 3 would make a total of 3(3-1) = 3 * 2 = 6. Each person registers 2 handshakes with the other 2 people in the group; 3 * 2.

Total Handshakes = n(n-1)

However, this includes each handshake twice (1 with 2, 2 with 1, 1 with 3, 3 with 1, 2 with 3 and 3 with 2) and since the orginal question wants to know how many different handshakes are possible we must divide by 2 to get the correct answer.

Total Different Handshakes = n(n-1)/2

Handshake Problem as a Combinations Problem

We can also solve this Handshake Problem as a combinations problem as C(n,2).

(objects) = number of people in the group
(sample) = 2, the number of people involved in each different handshake

The order of the items chosen in the subset does not matter so for a group of 3 it will count 1 with 2, 1 with 3, and 2 with 3 but ignore 2 with 1, 3 with 1, and 3 with 2 because these last 3 are duplicates of the first 3 respectively.

expanding the factorials,

cancelling and simplifying,

which is the same as the equation above.

Sandwich Combinations Problem

This is a classic math problem and asks something like How many sandwich combinations are possible? and this is how it generally goes.

Calculate the possible sandwich combinations if you can choose one item from each of the four categories:

  • 1 bread from 8 options
  • 1 meat from 5 options
  • 1 cheese from 5 options
  • 1 topping from 3 options

Often you will see the answer, without any reference to the combinations equation C(n,r), as the multiplication of the number possible options in each of the categories. In this case we calculate:

8 × 5 × 5 × 3 = 600
possible sandwich combinations

In terms of the combinations equation below, the number of possible options for each category is equal to the number of possible combinations for each category since we are only making 1 selection; for example C(8,1) = 8, C(5,1) = 5 and C(3,1) = 3 using the following equation:

C(n,r) = n! / ( r!(n – r)! )

We can use this combinations equation to calculate a more complex sandwich problem.

Sandwich Combinations Problem with Multiple Choices

Calculate the possible combinations if you can choose several items from each of the four categories:

  • 1 bread from 8 options
  • 3 meats from 5 options
  • 2 cheeses from 5 options
  • 0 to 3 toppings from 3 options

Applying the combinations equation, where order does not matter and replacements are not allowed, we calculate the number of possible combinations in each of the categories. You can use the calculator above to prove that each of these is true.

  • 1 bread from 8 options is C(8,1) = 8
  • 3 meats from 5 options C(5,3) = 10
  • 2 cheeses from 5 options C(5,2) = 10
  • 0 to 3 toppings from 3 options; we must calculate each possible number of choices from 0 to 3 and get C(3,0) + C(3,1) + C(3,2) + C(3,3) = 8

Multiplying the possible combinations for each category we calculate:

8 × 10 × 10 × 8 = 6,400
possible sandwich combinations

How many possible combinations are there if your customers are allowed to choose options like the following that still stay within the limits of the total number of portions allowed:

  • 2 portions of one meat and 1 portion of another?
  • 3 portions of one meat only?
  • 2 portions of one cheese only?

In the previous calculation, replacements were not allowed; customers had to choose 3 different meats and 2 different cheeses. Now replacements are allowed, customers can choose any item more than once when they select their portions. For meats and cheeses this is now a combinations replacement or multichoose problem using the combinations with replacements equation:

CR(n,r) = C(n+r-1, r) = (n+r-1)! / (r! (n – 1)!)

For meats, where the number of objects n = 5 and the number of choices r = 3, we can calculate either combinations replacement CR(5,3) = 35 or substitute terms and calculate combinations C(n+r-1, r) = C(5+3-1, 3) = C(7, 3) = 35.

Calculating cheese choices in the same way, we now have the total number of possible options for each category at

  • bread is 8
  • meat is 35
  • cheese is 15
  • toppings is 8

and finally we multiply to find the total

8 × 35 × 15 × 8 = 33,600
possible sandwich combinations!

How many combinations are possible if customers are also allowed replacements when choosing toppings?

Combinations as Selections

Suppose we have a set of 6 letters { A,B,C,D,E,F}. In how many ways can we select a group of 3 letters from this set? Suppose we find the number of arrangements of 3 letters possible from those 6 letters. That number would be 6P3. Consider the permutations that contain the letters A, B, and C. These are 3! = 6 ways, namely ABC, ACB, BAC, BCA, CAB, and CBA.

Now, what we want is the number of combinations and not the number of arrangements. In other words, the 6 permutations listed above would correspond to a single combination. Differently put, the order of things is not important; only the group/combination matters now in our selection. This means that the total number of combinations of 3 letters from the set of 6 letters available to us would be 6P3/3!

ways, since each combination is counted 3! times in the list of permutations. Thus, if we denote the number of combinations of 6 things taken 3 at a time by 6C3, we have: This is also said as 6 choose 3.

A few important results on combinations are as follows:

  • The number of ways of selecting n objects out of n objects is:
  • The number of ways of selecting 0 objects out of n objects is:

The number of ways of selecting 1 object out of n objects is:

Relationship Between Permutations and Combinations

Permutation and combination formulas and concepts have a lot of similarities. Suppose that you have n different objects. You have to determine the number of unique r-selections (selections that contain r objects) which can be made from this group of n objects. Think of a group of n people – you have to find the number of unique sub-groups of size r, which can be created from this group.

The number of permutations of size r will be nPr. In the list of nPr permutations, each unique selection will be counted r! times, because the objects in an r-selection can be permuted amongst themselves in r! ways. Thus, the number of unique combinations can be

Examples on Combinations

Example 1: Consider the word EDUCATION. This has 9 distinct letters. How many 3-letter permutations (words) can be formed using the letters of this word? We now know how to answer questions like this; the answer in this particular case will be 9P3

Consider the following 3-letter permutations formed using the letters A, E, T from EDUCATION:

AET , ATE, EAT,  ETA, TAE, TEA

These 6 different arrangements correspond to the same selection of letters, which is {A, E, T}. Thus, in the list of all 3-letter permutations, we will find that each unique Combination corresponds to 6 different arrangements. To find the number of unique 3-letter selections, we divide the number of 3-letter permutations by 6.

Hence, the number of 3-letter selections will be = 60480/ 6 = 10,080

Example 2: Out of a group of 5 people, a pair needs to be formed. The number of possible combinations can be calculated as follows.

Example 3:The number of 4-letter Combinations which can be made from the letters of the word DRIVEN is

Important Notes

  • Whenever you read the phrase “number of combinations”, think of the phrase “number of selections”. When you are selecting objects, the order of the objects does not matter. For example, XYZ and XZY are different arrangements but have the same selection.
  • The number of combinations of n distinct objects, taken r at a time (where r is less than n), is
  • This result above is derived from the fact that in the list of all permutations of size r, each unique selection is counted r! times.
  • Out of n objects, the number of ways of combinations 0 or n objects is 1; and the number of ways of selecting 1 object or (n – 1) object is n.
  • Out of n objects, the number of ways of selecting 2 objects is

Example 1. In a party of 10 people, each person shakes hands with every other person. How many possible combinations of handshakes can be made?

Solution:

Each unique handshake corresponds to a unique pair of persons. Also, note that the order of the two people in the pair does not matter. For example, if X and Y are two people, then XY and YX won’t be counted separately; only the pair {X, Y} will be counted. Here we consider the combination of any two people who shake their hand.

Thus, we need to find the number of ways in which 2 people can be selected out of 10. Clearly, the answer is:

Answer: 45 handshakes in total

Example 2. How many diagonals are there in a polygon with n sides?. (Use the concept of combinations to solve this.)

Solution:

If we select any two vertices of the polygon and join them, we will get either a diagonal or a side of the polygon. Here we can use the concept of combinations. The number of ways of selecting two vertices out of n is

Out of these selections, n correspond to the sides of the polygon. Thus, the number of diagonals is:

Example 3. A class has 25 students. For a school event, 10 students need to be chosen from this class. 4 of the students of the class decide that either four of them will participate in the event, or none of them will participate. What are the possible combination of 10 students?

Solution:

Given the specified constraint, we divide the set of all possible selections of 10 students into two groups:

  • The selections include the 4 students; we already have 4 students – we need to select 6 more students out of the remaining 21 students. This can be done in 21C6 ways.
  • Thus, the number of possible selections that include the 4 students is 21C6.
  • The selections do not include the 4 students; now we need to select 10 students out of the remaining 21. This can be done in 21C10 ways. Thus, the number of possible selections that do not include the 4 students is 21C10.

The total number of possible combinations under the specified constraint is 21C6+21C10.

Frequently Asked Questions

How To Apply Combinations Formula?

We calculate combinations using the combinations formula, and by using factorials and in terms of permutations. In general, suppose we have n things available to us, and we want to find the number of ways in which we can select r things out of these n things. We first find the number of all the permutations of these n things taken r at a time. That number would be nPr . Now, in this list of nPr

permutations, each combination will be counted r! times since r things can be permuted amongst themselves in r! ways. Thus, the total number of permutations and combinations of these n things, taken r at a time, denoted by nCr, will be:

What Are Combinations?

Combinations are selections made by taking some or all of a number of objects, irrespective of their arrangements. The number of combinations of n different things taken r at a time, denoted by nCr and it is given by,

where 0 ≤ r ≤ n. This forms the general combination formula which is nCr formula.

This formula to find the number of combinations by using r objects from the n objects, is also referred as the ncr formula.

What Is Combinations Formula?

The combinations formula is used to easily find the number of possible different groups of r objects each, which can be formed from the available n different objects. Combinations formula is the factorial of n, divided by the product of the factorial of r, and the factorial of the difference of n and r respectively.

The combinations formula is also referred to as ncr formula. To use the combinations formula we need to know the meaning of factorial, and we have n! = 1 × 2 × 3 × …. (n – 1) × n.

How many combinations of 4 items are there?

Unlike permutations, the order does not matter when computing combinations. So given 4 items a,b,c, and d, one has the following combination: abcd. As a combination, abcd is equivalent to cbad, which is equivalent to dbca, and so on. The answer changes should the 4 items come from a set with size greater than 4. For example, there are five ways to combine four items from a set of five items. If one wants to know how many permutations of 4 items are there, they need only compute 4! = 4 x 3 x 2 x 1 = 24, i.e., there are 24 permutations of 4 items.

What is the formulas for combinations and permutations?

The formula for the number of r-permutations of an n-set is P(n,r)=n!/(n-r)!, where ! denotes the factorial. The formula for the number of r-combinations of an n-set is C(n,r)=n!/r!(n-r)!=(P(n,r))/r!. We read C(n,r) as “n choose r.”

How do you find probability with combinations?

In order to compute the probability of a particular event occurring, simply divide the number of outcomes contained in the event by the total number of outcomes in the entire sample space. When outcomes are equally likely to take place, computing probabilities often amounts to nothing more than counting—which is not to say it is easy!

What is combination with example?

Fundamentally, combinations are unordered subsets of finite sets. Typically, a combination is the answer to a question that begins with “How many ways…” where order does not matter. For example, the number of ways three objects can be chosen from a set of five objects, without repetition, is ten. This answer can either be reached by counting all possible subsets of size three of a set of size five or by using the combination formula.

How do you find the total number of combinations?

To find the total number of combinations of size r from a set of size n, where r is less than or equal to n, use the combination formula: C(n,r)=n!/r!(n-r!)

This formula accounts for combinations without repetition, and a different formula is necessary to compute the total number of combinations with repetition.

What Are Combinations In Numbers?

Combinations are selections. Selecting r objects out of the given n objects is given by using the factorials. It is denoted by The combinations are the different subgroups that can be formed from the given larger group of objects.

How To Use The Combinations Formula?

Combinations are calculated using the combination formula while we need to choose r items out of n items. Here we use the factorial formula of n! = 1 × 2 × 3 ×…….(n – 1) × n.

Does order matter in combinations?

No, the order does not matter in combination. Just the number of selections or subgroups matters. The number of dresses in the wardrobe can be selected at random in order. Picking 2 clothes out of 8 from the wardrobe requires ways. Here we consider the set of two and do not look into the order of the selection.

What are the possible combinations of 6 numbers?

The possible combinations (selections) out of 6 different numbers are as follows:

What are the possible combinations out of the digits 1234?

The total number of combinations are possible this way: choosing 1 digit out of 4, choosing 2 digits out of 4, choosing 3 digits out of four and choosing 4 digits out of 4 =

How Are Permutations and Combinations Related?

The permutations and combinations are related using the combination formula

The combinations are the selection of r things taken from n different things, and permutation is the different arrangement of those r things.

What Are The Differences Between Permutations and Combinations?

Permutations are seen as arrangements of r things out of n things, whereas combinations are seen as selections of r things out of n things.

For the given r things out of n things, the number of permutations are greater than the number of combinations.

What Are the Combination Examples?

The combination examples include the groups formed from dissimilar obects.The formation of a committee, the sport team, set of different stationary objects, team of people are some of the combination examples.